Question: The following equation is true for all real values of $x$ for which the expression on the left is defined, and $B$ is a polynomial expression. $\dfrac{B}{2x^3+14x^2} \div \dfrac{5x-35}{10x^2-490}=1$ What is $B$ ? $B=$
Explanation: The left side of the equation is a quotient of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting quotient on the left side should cancel out completely. In order to solve for $B$, let's divide the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The denominator, $2x^3+14x^2$, of the dividend can be factored as $2x^2(x+7)$ by factoring out $2x^2$. The numerator, $5x-35$, of the divisor can be factored as $5(x-7)$ by factoring out $5$. The denominator, $10x^2-490$, of the divisor can be factored as $10(x+7)(x-7)$ by factoring out $10$ and using the difference of squares pattern. Now the quotient looks as follows: $\dfrac{B}{2x^2(x+7)} \div \dfrac{5(x-7)}{10(x+7)(x-7)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{B}{2x^2(x+7)} \div \dfrac{5(x-7)}{10(x+7)(x-7)} \\\\\\ &= \dfrac{B}{2x^2(x+7)} \cdot \dfrac{10(x+7)(x-7)}{5(x-7)} &\text{Flip the divisor} \\\\\\ &= \dfrac{B \cdot 10(x+7)(x-7)}{2x^2(x+7) \cdot 5(x-7)} &\text{Multiply across.}\\\\\\ &= \dfrac{B {\cancel{2}}\cdot{\cancel{5}}{\cancel{(x+7)}}{\cancel{(x-7)}}}{{\cancel{2}}x^2{\cancel{(x+7)}} {\cancel{5}}{\cancel{(x-7)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{B}{x^2} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{B}{x^2}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $B=x^2$.